Question: In which situation would a space probe most likely experience centripetal acceleration?
as it revolves around a planet
as it flies straight past a moon
as it is pulled in a line toward the Sun
as it lifts off from Earth
Answer:
When "space probe revolves around a planet" most likely to experience centripetal acceleration
Explanation:
Centripetal acceleration defined as the rate in change of tangential velocity. Also, as per Newton's second law, any kind of force will be directly proportional to the acceleration attained by the object. So, for centripetal acceleration, the force will be the centripetal force. The centripetal force will be acting on an object rotating in a circular motion with its direction of force towards the center. Thus, centripetal acceleration will be experienced by an object or a space probe when it is in a circular motion. It means the space probe is revolving around a planet.
90 degrees - 30 = 60 degrees
Velocity = (5m/s - 4.35m/s x cos(30)) / cos(60)
Velocity = 2.47 m/s
The answer is D) 2.47 m/s at 61.9 degrees
By definition, the refractive index is
n = c/v
where c = 3 x 10⁸ m/s, the speed of light in vacuum
v = the speed of light in the medium (the liquid).
The frequency of the light source is
f = (3 x 10⁸ m/s)/(495 x 10⁻⁹ m) = 6.0606 x 10¹⁴ Hz
Because the wavelength in the liquid is 434 nm = 434 x 10⁻⁹ m,
v = (6.0606 x 10¹⁴ 1/s)*(434 x 10⁻⁹ m) = 2.6303 x 10⁸ m/s
The refractive index is (3 x 10⁸)/(2.6303 x 10⁸) = 1.1406
Answer: a. 1.14
Um where are the options for the answers?
Answer:
Speed of the aircraft = 36.64 m/s
Explanation:
Consider the vertical motion of the projectile,
We have equation of motion s = ut+0.5at²
Let the velocity of plane be v.
Vertical velocity = vsin55 = Initial velocity of projectile in vertical direction = u
acceleration, a = 9.81 m/s²
displacement , s = 554 m
time, t = 8 s
Substituting,
554 = vsin55 x 8 +0.5 x 9.81 x 8²
v = 36.64 m/s
Speed of the aircraft = 36.64 m/s
